COMEDK · Chemistry · 7. Chemical Equilibrium
In the equilibrium, \(2 \mathrm{NH}_3 \rightleftharpoons \mathrm{~N}_2+3 \mathrm{H}_2\), 6 mol of \(\mathrm{NH}_3\) is taken in 10 L flask. If concentration of \(\mathrm{N}_2\) at equilibrium is \(x\), then concentration of \(\mathrm{NH}_3\) at equilibrium is
- A \(0.6-x\)
- B \(0.6-2 x\)
- C \(0.6-\frac{x}{2}\)
- D \(0.6-\frac{x}{3}\)
Answer & Solution
Correct Answer
(B) \(0.6-2 x\)
Step-by-step Solution
Detailed explanation
\begin{array}{lccc} & 2 \mathrm{NH}_3 \rightleftharpoons & \mathrm{~N}_2+3 \mathrm{H}_2 \\ \text{Initial} & \frac{6}{\mathrm{~mol}} \mathrm{~mol} & 0 & 0 \\ \text{concentration} & 10 & & \\ & =0.6 \mathrm{~mol} / \mathrm{L} & & \\ \begin{array}{l} \text{Concentration} \\…
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